To solve the problems on this worksheet and verify the solutions provided, I will work through each of the problems step by step. Here’s the breakdown:

Problem S:

$\frac{4}{9} = \frac{x}{15}$

Cross multiply: $4 \cdot 15 = 9 \cdot x$ $60 = 9x$ $x = \frac{60}{9}$ Simplify: $x = 6.\overline{6}$ or approximately $x = 6.7$.

Problem I:

$\frac{5}{16} = \frac{3}{a}$

Cross multiply: $5 \cdot a = 16 \cdot 3$ $5a = 48$ $a = \frac{48}{5}$ Simplify: $a = 9.6$.

Problem G:

$\frac{-2}{11} = \frac{b}{50}$

Cross multiply: $-2 \cdot 50 = 11 \cdot b$ $-100 = 11b$ $b = \frac{-100}{11}$ Approximate: $b = -9.1$.

Problem U:

$\frac{k}{5} = \frac{12}{-7}$

Cross multiply: $k \cdot -7 = 12 \cdot 5$ $-7k = 60$ $k = \frac{-60}{7}$ Simplify: $k = -8.57$ (rounded).

Problem E:

$\frac{27}{m} = \frac{10}{3}$

Cross multiply: $27 \cdot 3 = 10 \cdot m$ $81 = 10m$ $m = \frac{81}{10}$ Simplify: $m = 8.1$.

Problem L:

$\frac{7}{4} = \frac{2x}{13}$

Cross multiply: $7 \cdot 13 = 4 \cdot 2x$ $91 = 8x$ $x = \frac{91}{8}$ Simplify: $x = 11.375$ or $x \approx 11.4$.

Problem K:

$\frac{1}{6y} = \frac{5}{33}$

Cross multiply: $1 \cdot 33 = 6y \cdot 5$ $33 = 30y$ $y = \frac{33}{30}$ Simplify: $y = 1.1$.

Problem C:

$\frac{t - 3}{2} = \frac{8}{9}$

Cross multiply: $(t - 3) \cdot 9 = 2 \cdot 8$ $9t - 27 = 16$ $9t = 43$ $t = \frac{43}{9}$ Simplify: $t = 4.\overline{7}$ or $t \approx 4.778$.

Problem N:

$\frac{8 + a}{15} = \frac{1 + a}{18}$

Cross multiply: $(8 + a) \cdot 18 = (1 + a) \cdot 15$ $144 + 18a = 15 + 15a$ $3a = -129$ $a = -43$.

Problem B:

$\frac{7 - 2n}{7 + 2n} = \frac{1}{7}$

Cross multiply: $(7 - 2n) \cdot 7 = (7 + 2n) \cdot 1$ $49 - 14n = 7 + 2n$ $42 = 16n$ $n = \frac{42}{16}$ Simplify: $n = 2.625$.

Problem A:

$\frac{11b}{6} = b - 5$

Multiply through by 6 to eliminate the fraction: $11b = 6(b - 5)$ $11b = 6b - 30$ $5b = -30$ $b = -6$.

Problem T:

Argyle reads 7 pages in 5 minutes. How many pages in 2 hours?

Convert 2 hours to minutes: $2 \text{ hours} = 120 \text{ minutes}$.

Set up proportion: $\frac{7}{5} = \frac{x}{120}$ Cross multiply: $7 \cdot 120 = 5 \cdot x$ $840 = 5x$ $x = \frac{840}{5}$ $x = 168$.

Problem X:

There are 1200 calories in one pound of ground beef. How many calories are in 3 ounces?

Convert 3 ounces to a fraction of a pound: $3 \text{ oz} = \frac{3}{16} \text{ lb}$.

Set up proportion: $\frac{1200}{1} = \frac{x}{\frac{3}{16}}$ Multiply: $x = 1200 \cdot \frac{3}{16}$ $x = 225$.

Would you like further explanations or clarifications? Here are some relative questions for exploration:

1. What are the key steps to solve proportion problems systematically?
2. How does cross-multiplication work and why is it valid in equations involving fractions?
3. How can you check the solutions of these problems for accuracy?
4. What are practical applications of proportions in real life?
5. What are some common pitfalls to avoid when solving equations like these?

Tip: Always simplify fractions at every step to make calculations easier and avoid errors!